Question: Suppose we have a vector field $f(x, y) = (e^x, y^{-1})$ and a curve $C$ that is parameterized by $\alpha(t) = \left( 3t, t^2 \right)$ for $1 < t < 2$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Solution: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (e^x, y^{-1})$ and $\alpha(t) = (3t, t^2)$. $\begin{aligned} &f(\alpha(t)) = \left( e^{3t}, t^{-2} \right) \\ \\ &\alpha'(t) = (3, 2t) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_1^2 \left( e^{3t}, t^{-2} \right) \cdot (3, 2t) \, dt$ Let's solve the integral. $\begin{aligned} &\int_1^2 \left( e^{3t}, t^{-2} \right) \cdot (3, 2t) \, dt \\ \\ &= \int_1^2 3e^{3t} + 2t^{-1} \, dt \\ \\ &= \left[ e^{3t} + 2 \ln(t) \right]_1^2 \\ \\ &= (e^6 + 2 \ln(2)) - (e^3 + 2 \ln(1)) \\ \\ &= e^6 - e^3 + 2\ln(2) \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = e^6 - e^3 + 2\ln(2)$.